Ємність буферного розчину та діапазон pH

As we've been studying acids and bases, we've started investigating buffers. They can resist changes in pH. Making a buffer required mixing a weak acid with its conjugate base. So typically, we start with a weak acid and add the ionic salt of its conjugate base. We're taking advantage of the common ion effect! The acid generates hydronium and neutralizes any added base. The conjugate base will neutralize any added acid. Today, we'll explore buffers in more detail.

Calculating a Buffer's pH

First, let's recall how to calculate a buffer's pH. The easiest way is to use the Henderson-Hasselbalch equation. Let's look at an example.

What is the pH of a buffer made from 0.23 molar acetic acid and 0.14 molar sodium acetate? Acetic acid's dissociation constant value is 4.75.

$pH=pK_{a}+ \log{\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}}\rightarrow pH=4.75+\log{\frac{0.14}{0.23}}=4.53$.

Notice that the pH is slightly lower than the $pK_{a}$ because we have more acid than base. Are you wondering why we can use the initial concentrations of acetic acid and acetate? We could use an ICE table and calculate a new equilibrium concentration. But, notice the pH. Our hydrogen ion concentration is tiny. So, the acetic acid and acetate concentrations don't change much. So, we assume they're constant.

Determining a Buffer's pH Range

Imagine you walked into the lab. Your teacher tells you that you'll be performing an experiment. But, it requires a consistent pH. What would you do? We're studying buffers. So, you'd probably know to use one. But, which would you use?

All buffers have a pH range. The pH range measures the span over which a buffer solution can regulate pH. It only depends on the compounds used to make it. So it's an intensive property. In the lab, we label their pH range. One example is citric acid/citrate. It is a standard buffer in analytical chemistry. If you picked up a bottle of citric acid buffer, you'd read that its pH range is 3.0-6.2. So, if we are within that range, the buffer solution is effective. The change in pH will be tiny. If we are outside that range, the buffer is ineffective. So, the pH will change dramatically.

So, how do we calculate the pH range? We know a buffer's pH depends on the ratio $\frac{[Conjugate]}{[Acid/base]}$. So, what proportion gives us a usable buffer? Thankfully, chemists have found experimentally that most buffers are $0.1 \leq \frac{[Conjugate]}{[Acid/base]} \leq 10$. Let's put these numbers into the Henderson-Hasselbalch equation!

$pH= pK_{a} + \log{(\frac{[Conjugate]}{[Acid/base]})}$

For $\frac{[Conjugate]}{[Acid/base]}=10$...

$pH= pK_{a} + \log{(10)}=pK_{a}+1$

For $\frac{[Conjugate]}{[Acid/base]}=0.1$...

$pH= pK_{a} + \log{(0.1)}=pK_{a}-1$

So, a buffer's pH range is $pK_{a} \pm 1$. Technically, this is an approximation. The experimental values come from titration experiments. But, it's a useful rule of thumb.

Example Problem: Acetic Acid

What is the pH range of an acetic acid/acetate buffer?

The first step is to look up the $pK_{a}$ value. Acetic acid is a standard buffer. So, you should probably memorize that its $pK_{a}=4.75$. So, the pH range is $3.75-5.75$. When we look at the experimental value, it's $3.7-5.6$! So, our approximation was unbelievably close!

Calculating Buffer Capacity

Buffers also have a capacity. The buffer capacity is the amount of acid/base the buffer can consume. It measures its ability to keep the pH within the pH range. The buffer's ability to absorb acids or bases is an extensive property. It depends on the initial concentrations of the acid/base and its conjugate.

Let's look at how we calculate buffer capacity. We denote it with the greek letter beta ($\beta$). The buffer wants to stay within its pH range. So, it has to neutralize any added acid/base. Therefore, we measure the moles per liter of neutralized acid/base. So, our formula is $\beta= \frac{\Delta [acid/base/conjugate]}{\Delta pH}$. The units are a little strange. It's the concentration change per unit of pH change. For example, let's say we added strong acid to a buffer. We calculated our pH capacity as $0.203 \frac{\frac{mols}{L}}{pH}$. So, for every one unit of pH change, our base concentration changed by $0.203\frac{mols}{L}$.

Example Problem: Acetic Acid Again

Let's look at an acetic acid buffer. The initial concentrations of acetic acid and acetate are equal at 0.15 molar. Calculate the buffer capacity across the pH range. Assume we are adding a strong acid and base.

Start by finding your initial pH with the Henderson-Haselbalch equation.

$pH=pK_{a} + \log{(\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]})} \rightarrow pH= pka + \log{(1)} \rightarrow pH=pKa=4.75$.

We reach the lower end of the pH range by adding a strong acid. The acid neutralizes the acetate ions. We also generate acetic acid.

$\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]} \rightarrow \frac{[CH_{3}COO^{-}]-x}{[CH_{3}COOH]+x}=0.1$.

So, we can solve for x.

$\frac{0.15-x}{0.15+x}=0.1 \rightarrow .015+0.1x=0.15-x \rightarrow 1.1x= 0.135 \rightarrow x=0.12 \frac{moles}{L}\, of\, H^{+}$.

We can do the same procedure for the upper limit. But, we reach the upper end by adding a strong base. The base neutralizes the acetic acid. So, we acetate ions.

$\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]} \rightarrow \frac{[CH_{3}COO^{-}]+x}{[CH_{3}COOH]-x}=10$.

Again, we solve for x.

$\frac{0.15+x}{0.15-x}=10.0 \rightarrow 1.5-10x=0.15+x \rightarrow 11x= 1.35 \rightarrow x=0.12 \frac{moles}{L}\, of\, OH^{-}$.

Why are the two answers identical? Because we started with equal concentrations of acetic acid and acetate. What if we had different initial concentrations? Our procedure would be the same. But, the "x" values would differ.

Now, let's calculate the buffer capacity.

$\beta=\frac{0.122}{5.75-4.75}= 0.12 \frac{\frac{moles}{L}}{pH}$.

Example 2: Limited pH

For our second example, let's narrow the pH range. We are doing an experiment that needs a pH below 5.40. For ease, let's say we have an acetic acid buffer. The initial concentration of acetate was 0.12 molar. The initial concentration of acetic acid was 0.14 molar. What is the buffer capacity for our experiment?

For this problem, we need to use the whole Henderson-Hasselbalch equation twice. The first time, we'll calculate our initial pH. The second time, we'll solve for the concentration change at our pH limit. Then, we'll calculate the buffer capacity.

$pH_{o}= pK_{a} + \log{(\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]})} = 4.75 + \log{(\frac{0.12}{0.14})}=4.68$

Now we can solve for concentration change. We reach our upper limit by adding a strong base. So, we consume acetic acid and generate acetate.

$pH= pK_{a} + \log{(\frac{[CH_{3}COO^{-}]+x}{[CH_{3}COOH]-x})}$

$5.40= 4.75 + \log{(\frac{0.12+x}{0.14-x})} \rightarrow 10^{5.40-4.75}=\frac{0.12+x}{0.14-x} \rightarrow 5.24(0.14-x)=0.12+x \rightarrow x=0.098\frac{mols}{L}$.

Finally, we can calculate our buffer capacity.

$\beta= \frac{x}{\Delta pH}=\frac{0.098}{5.24-4.68}=0.175\frac{\frac{mols}{L}}{pH}$

For Further Study

The buffer capacities and pH ranges have many different applications. If you want to learn more, check out these other explanations.

• Acid-base indicators

• Selecting an indicator

• Titration of a weak base with a strong acid

• Titration of a weak acid with a strong base

• Titrations of polyprotic acids

(Video) What Is Buffer Capacity?

by Sarah Chem

This short video is a great explanation of buffer capacity!

(Video) Buffer Range Calculations

=== by StudyForce

In this video, StudyForce demonstrates two pH range example problems. In both, you have to use the Henderson-Hasselbalch equation. If you're struggling with these types of problems, this video can help!